The figure Figure 1 shows a small plant near a thin lens The

The figure (Figure 1) shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 2.00 cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions.

Part B

What is the focal length of the lens?

Express your answer using two significant figures.

Part C

Calculate where the image should be.

Solution

Part A

The lens is diverging , therefore the rays which become parallel to optical axis must be moving along a direction which meets to the focus ( in real or virtual way)

Thus focal length =3×2cm = 6 cm

Part B

f = -6 cm

u= -6cm

V = uf/(u+f) = -6*-6/(-12) = -3 cm